如何存储整数值等于10 ^ 18在C程序或C ++? [英] How to store a integer value equal to 10^18 in c programs or c++?
问题描述
我怎么可以存储在C变大的整数值?
如果我声明与 int类型的;
将无法正常工作。
我与使用该得到long long int
。它是行不通的。
如果(A> = 0&放大器;&安培; A< =(1000000000000000000))
什么来声明一个变量,这样就不会那么任何error.It应该是整数。
编译器错误
整型常量是long类型太大。
让我们看看你写了这个如果
语句:
如果(A> = 0&放大器;&安培; A< =(1000000000000000000))
1000000000000000000
太大为一体的文字,所以你需要一个更大的文字类型。你应该申报 A
为的int64_t
和不喜欢该水平的研究:
如果(A> = INT64_C(0)及和放大器; A< = INT64_C(1000000000000000000))
请注意,这将只在 C99
或 C ++ 11
编译器,当你<$ C $工作C>的#include&LT; cstdint&GT; 或的#include&LT; stdint.h&GT;
编辑:在目前的标准草案,你可以找到这个句子(2.14.2 / 2):
类型字面的整数的是第一相应列表的
表6在其值可以被重新presented
块引用>这意味着,编译器会自动使用所需的文本类型,使您的文字配合。顺便说一句,我没有看到那种编译器。
How can I store a large integer value in a variable of C ? If i am declaring a with
int a;
it won't work. I have used this withlong long int
.It is not working.if( a>=0 && a <= (1000000000000000000))
What to declare variable a so that it will not so any error.It should be integer.
Compiler error integer constant is too large for long type.
解决方案Lets look at this
if
statement you wrote:if( a>=0 && a <= (1000000000000000000))
1000000000000000000
is too big for an integral literal so you will need a bigger literal type. You should declarea
as anint64_t
and do the comparion like that:if( a>=INT64_C(0) && a <= INT64_C(1000000000000000000))
Note that this will only work in a
C99
orC++11
compiler when you#include <cstdint>
or#include <stdint.h>
Edit: in current draft of the standard you can find this sentence (2.14.2/2):
The type of an integer literal is the first of the corresponding list in Table 6 in which its value can be represented.
It means that compiler should use the required literal type automatically to make your literal fit. Btw I didn't see that kind of compiler.
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